A puck of mass 0.60 kg approaches a second, identical puck that is stationary on frictionless ice. The initial speed of the moving puck is 3.0 m/s. After the collision, one puck leaves with a speed v1 at 30° to the original line of motion. The second puck leaves with speed v2 at 60°.
Find v1 and v2
Thank you!!
Find v1 and v2
Thank you!!
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The problem only makes sense if the pucks are going in different directions. Since the momentum is conserved the speeds can be calculated by analyzing the components of the momentum perpendicular to the original movement.
Calling the axes along the original pucks movement x and the perpendicular y.
Assuming that the to pucks moves in different directions along the y axes.
The mass are both called m.
So the total momentum along the x axes should be (v1x*m + v2x*m) = 3.0 * m => v1x + v2x = 3.0.
Along the y-axes the total momentum is zero before the collision and the same after so v1y*m - v2y*m = 0 => v1y = v2y (We subtract v2y since it is going in the opposite direction.
To find v1x, v1y, v2x and v2y we use sin and cos.
sin(30) = v1y/v1, sin(60) = v2y/v2, cos(30) = v1x/v1 and cos(60) = v2x/v2.
Since sin(30) = cos(60) = 0.5 and cos(30) = sin(60) lets simplify a little:
0.5 = v1y/v1, sin(60) = v2y/v2, sin(60) = v1x/v1 and 0.5 = v2x/v2
v1y = v2y => 0.5*v1 = sin(60)*v2 => (0.5*v1)/sin(60) = v2
v1x + v2x = 3.0 => sin(60)*v1 + 0.5*v2 = 3.0 => sin(60)*v1 + 0.5*(0.5*v1)/sin(60) = 3 =>
(sin(60) + 0.25/sin(60))*v1 = 3 => v1 = 2.6
v2 = (0.5*v1)/sin(60) = (0.5*2.6)/sin(60) = 1.5
(a negative velocity makes no physical sense)
Calling the axes along the original pucks movement x and the perpendicular y.
Assuming that the to pucks moves in different directions along the y axes.
The mass are both called m.
So the total momentum along the x axes should be (v1x*m + v2x*m) = 3.0 * m => v1x + v2x = 3.0.
Along the y-axes the total momentum is zero before the collision and the same after so v1y*m - v2y*m = 0 => v1y = v2y (We subtract v2y since it is going in the opposite direction.
To find v1x, v1y, v2x and v2y we use sin and cos.
sin(30) = v1y/v1, sin(60) = v2y/v2, cos(30) = v1x/v1 and cos(60) = v2x/v2.
Since sin(30) = cos(60) = 0.5 and cos(30) = sin(60) lets simplify a little:
0.5 = v1y/v1, sin(60) = v2y/v2, sin(60) = v1x/v1 and 0.5 = v2x/v2
v1y = v2y => 0.5*v1 = sin(60)*v2 => (0.5*v1)/sin(60) = v2
v1x + v2x = 3.0 => sin(60)*v1 + 0.5*v2 = 3.0 => sin(60)*v1 + 0.5*(0.5*v1)/sin(60) = 3 =>
(sin(60) + 0.25/sin(60))*v1 = 3 => v1 = 2.6
v2 = (0.5*v1)/sin(60) = (0.5*2.6)/sin(60) = 1.5
(a negative velocity makes no physical sense)
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equal masses So m1=m2= m=0.6 kg
Initial momentum = 0.60 kg *3m/s = 1.8 kg m/s
X direction momentum of m1= 0.6 * v1 cos 30 = 0.52 v1 kg m/s
X direction momentum of m2 = 0.6v2 cos 60 = 0.3v 2 kg m/s
Y direction momentum of m1 = v1 sin 30 = 0.3 v1 kg m/s
Y direction momentum of m2 = v2 sin 30 = 0.52 v2 kg m/s
final momentum of both the masses = their initial momentum before collision
0.52v1+ 0.6v2 =1.8 along X direction
0.6 v1+0.52 v2 = 0 along Y direction
v1= -0.52 / 0.6 v2 = -0.87 v2
substitute this in x direction equation
(-0.87v2)+0.52 v2 = 1.8
- 0.35 v2 = 1.8 => V2 = - 5.19 m/s
V1 = -0.87* (-0.5.19) = 4.52m/s
Initial momentum = 0.60 kg *3m/s = 1.8 kg m/s
X direction momentum of m1= 0.6 * v1 cos 30 = 0.52 v1 kg m/s
X direction momentum of m2 = 0.6v2 cos 60 = 0.3v 2 kg m/s
Y direction momentum of m1 = v1 sin 30 = 0.3 v1 kg m/s
Y direction momentum of m2 = v2 sin 30 = 0.52 v2 kg m/s
final momentum of both the masses = their initial momentum before collision
0.52v1+ 0.6v2 =1.8 along X direction
0.6 v1+0.52 v2 = 0 along Y direction
v1= -0.52 / 0.6 v2 = -0.87 v2
substitute this in x direction equation
(-0.87v2)+0.52 v2 = 1.8
- 0.35 v2 = 1.8 => V2 = - 5.19 m/s
V1 = -0.87* (-0.5.19) = 4.52m/s